Quiz10Solution - CE 3700 Fluid Mechanics Homework Quiz 10...

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Unformatted text preview: CE 3700 Fluid Mechanics Homework Quiz 10 12/9/05 Name: 5 0| UL‘l'T'Ls‘A SHOW ALL WORK AND ALL UNITS. Ifyou use the Moody Diagramz draw your lines on the diagram so the grader can (allow your aggroach. 1. a) Find QA and Q3. b) Find th and hm. (Ignore minor losses such as bends, entrance/exit losses, valves, etc.) Pipe A, QA, th ——-> PipeA: lSOOftlong,D=10in, f= 0.018 Pipe B: 900 ft long; D = 8 in., f= 0.017 Q—>=6fl3/S —’ Q=6fl3/s I . PipeB,QB,hIB hi}; 1 “$3 Ea v7. V'L. ‘FA VA 1% '3 “$3 05 1% QA L Q”- «CA L“ 2 — E ,E g ._ DA (Vivi) % 95 $95.97. M L i = a; (Eco L) Twoefiua’lms A DE Twmnlanowvxs QA + QB 2 6:0}; (EQ Z) \ _ & L D: WHMWWWWWW E62 1 a QR " {1A Ti W G? AA ; $(1T%n)2_2 015%g’uz A I ,8. 7— _ L Subsirzl'uhnj EQL iuh 5&1 ‘- 13: 4- 173?”) M m “LE [2'1 OAS : 3 illeLABg- Qg 4" Q3 (:5 Q Jr GK 2 age \y’ 0.018 13000 ( 5m 3 L'____W__,,«—J \.7_oc> 7. 7. ln ‘ ln = lit iA‘ _ 1800” {6.00 ms} M - £12.. H on 23 — 0.0‘8 lToiu lama/51) :. 1L7 P _-...___._.., 2. For the cross section shown below, find the total discharge. The Manning’s 11 value for the left and right overbanks is 0.040. The Manning’s 11 value for the channel is 0.016. The bed slope and energy grade line are the same and are equal to a 3-ft drop over 1 mile. C \r + Lag- lmwmnel ovligbwnk 2509 17.9- 200“ 12.“ 2503+ 3 Dr Channel n: 0.0“, 50 = 930” = 0,0005% A : (zoom: and + (JileflxIZP-LXl) + Qon \w-XL): 4780“; LII-M L—V—‘__———2 l————-‘(——Q——v “Hoe n1 I‘H at 140 “L P = Zoom 4— z (Wm-H— aW’ : 2233 9* W lea7¢L A , 47 007' _ Rh: 7‘ ' 233319 “ ZO'LHR V 1/ Y2 1" Q = A = L“ Memoma e channel 0‘0"” W gr“) 007.15 7-979 : 1%,200 9% W H : 0040 SO = 0.0005725 A 2 (L309 x to“) + (fix to“ we“) '5 ZSSD 347’ P 2 EMF—e W = lemma W N.qu Rh: 1% =M= mama 1w»: F1 V7. 1/2 Q m ASIIIK? e ‘5 @fiofi‘llofimin“) bVEl’lannL _ n ° 0.040 : [0, 2w 03/; “Raj : 717/005? + l<lo,’Leom/S) 2 (11120“? CE 3700 Fluid Mechanics Homework Quiz 10 12/9/05 Name: 3 O lu‘H 0 A SHOW ALL WORK AND ALL UNITS. 5 It you use the Moody DiagramI draw your lines on the diagram so the grader can [bllow your gggmach. l. a) Find QA and QB. b) Find th and hm. (Ignore minor losses such as bends, entrance/exit losses, valves, etc.) Pipe A, QA: th ——> Pipe A: 2000 ft long, D = 1.000 R, f= 0.016 Pipe B: 1000 fi long; D = 0.667 f1, f= 0.015 Q=8ft3/s —> Q=8fi3/s PipeBsQBahiE L‘QA = “$13 a % i4: _ stats 2:: A 7.3 ‘03 L3 L L mewm ( 1A 6:5" ‘9'?“ = i3 5% QB (EQ 1) E Tw unknown; 0 QA + Qg 2 (Bi—*3 (an) r—"fi E ——9 Q — 3. PA: Q Q1 A V ’gA LA 935 ‘2 Subs’nhfiwc‘ E62 1 m’rb EQZ 5 11% L3 1) s i; is 6:9 QB + Q3 : 8 0.015- loooH (1009‘); Q + Q 0‘0 to, Zooo R 0.0.7134 G E \ __.__._.__..n,(_,,,v~.__.. ._J LEM- h \A LA v7” 200041 WW 1 2 1: a —— A 2 I ‘0 f.—— I 3 «CA 593 11; DA 23 o 0\ Wu 43“ W51) 2. For the cross section shown below, find the total discharge. The Manning’s n value for the left and right overbanks is 0.045. The Manning’s n value for the channel is 0.018. The bed slope and energy grade line are the same and are equal to a 3—ft drop over 1 mile. KC k+ ovLeevfitw MACH/Md Dveglazzvgk 45° 3% M_ Vl: 0.0l8 So: : 0,000‘3é5 5280 A 2 ((400 x28”) + (M4 x80 xiXL) + llsflx WM?) 2 #635 01 L’._ov_..___2 W___V__w’__4l gene 1'25 3% P: Moll +ZJQSJ€+Y+QS£QD = $2.4” 9w) Z\.Z| g 2 is: , HSBS‘HL _ Queue” l" P [MA—l £4 “ ' 4 1/ H ‘41 V1 3/3 __ LJ 1 3 — .‘_' 1 0.000968 M56, (90¢:de _ '0 ASD KL‘ _ o_o\3 6535“)( ) < = 7¢,z\o “73 Oveflobmb H : So: 0,000g68 A = (175Hxlsfl)+ (liwzuxmfl 2 234,0 W "—1 P : I’I‘SH + J(Igfl)L—LQ3 )1 ~ : H33 45;, W'Td “3,35 Rh : 1;") 2 M: 2. ‘L.Lb 013.“; ‘ l/L 743 L44 "2 1’3 _ m Qoverbkxk : I; A SD Rh - o‘cqg (13Q0£L1)(0'000S(08)(ll'zo‘pl) 2 ‘1570 9‘?! Tani :4 7912\0J33} 4— 2087000475) f” $376130 “3% CE 3700 Fluid Mechanics Homework Quiz 10 12/9/05 Name: SOlWi—i o n SHOW ALL WORK AND ALL UNITS. It you use the Moody Diagramx draw your lines on the diagram so the gutter can [allow your aggroach. 1. 3) Find QA and Q3. b) Find th and hug. (Ignore minor losses such as bends, entrance/exit losses, valves, etc.) A: QA: th —> Pipe A: 2500 ft long, D = 1.25 ft, f= 0.018 Pipe B: 1200 ft long; D =1.00 ft, f= 0.017 0= 10 fiS/s —> Q= 10 fts/s —’ _> Pipe B! QB: hEB A $6 1 '1— l1: yfi : 4C lie— Us A DA 23 B De 7'3 L Q?— L JFA r— Af _ L; Q P 1 1 ' “C8 “ii—75'.— “on 54 Us @5va % LA L _ LL 1 TWO Piual'iov‘s ‘pA D: QA — “:5 DE Q3 (EQ 1) Two unknowns QA 4 6% r \o 2‘5 (“EC”) (4’1 H L10? EQi—a 62A: iii’Ifst B Sula Sl’H‘Ldej EQ i 'van EQ Z t a 2:2 £1 E?- QAG‘ Qg 4’ QB : lb 5 {LA LA 193 P” -l _ - om .Zgga Q3 + Qg ' ms a 0.0m 25mm 1.009 1.116 7. LA v 1 2am? Mil, “H = We = JEA fix 5? t 0918 “15” 43”““0 :. norm ¥+ ’__—§ 2. For the cross section shown below, find the total discharge. The Manning’s 11 value for the left and right overbanks is 0.042. The Manning’s n value for the channel is 0.017. The bed slope and energy grade line are the same and are equal to a 4-ft drop over 1 mile. Leg— ' a “ M” overload; ' C(A “flak ofggbank {50 H 20% no H 7.094 “Jon M l 4 LL Wi— n: 0,017 So : SLED : 0,00075‘8 1 Ar : QM}. x 332+) 4». (4i). mom x 743434509 J, (2M4 x 183*)(1): S300lll ?: HOE! +2@ : mags} > 18.28 a“: 15$ = ifi‘f—l; = 3L8le = 2:16.: = wil<aowmllu :- luslflroo 19“}? W m: O'o4L Soy: 0‘00013‘5 Afi: (157)11¥\B£+\+ (fixmflxlsfiv 2 L862 .5341 "P 2 \Soer + 1 l QBH)Z+(\B?l)L : \‘l'Sfi‘fl W 7’ ‘2:'&‘:—¥————‘LBGLH ’2 1 l9? “ P naeu ("3 A V 1/3 \4‘1 V; 1’3 _ l\ 1 __ I overhaul; " “g A 35 KM —- WL (134207’>(0.0007§3)( $3”) 2: 17‘180 “3/; Talml : l1??le Li; 4. 1(17/‘180‘m/s) z : {£34,360 93/; ...
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This note was uploaded on 04/03/2008 for the course CE CE 3700 taught by Professor Peyton during the Fall '05 term at Missouri (Mizzou).

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Quiz10Solution - CE 3700 Fluid Mechanics Homework Quiz 10...

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